3
$\begingroup$

Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.

First some notational specifications: Let us define $\tau$ as the topology of $X$ $\tau_{Y}=\{U \cap Y; U \in \tau \}$; $\tau_{A}^Y=\{V \cap A; V \in \tau_{Y} \}$ and $\tau_{A}^X=\{U \cap A; U \in \tau \}$

Let us show that $\tau_{A}^Y=\tau_{A}^X$

$\subset$: Let $V \cap A$ be an element of $\tau_{A}^Y$ such that $V \in \tau_{Y}$

Then, there exists $U\in \tau$ such that $V=U\cap Y$

$V\cap A=(U\cap Y)\cap A=U\cap (Y\cap A)= U\cap A$

Hence: $V\cap A \in \tau_{A}^X$ and so: $\tau_{A}^Y \subset \tau_{A}^X$

$\supset$: Let $U \cap A \in \tau_{A}^X$

$U\in \tau \iff U \text{ open in } X$

But $U \cap Y$ open in $Y \iff U \cap Y \in \tau_{Y}$

Hence $(U \cap Y) \cap A \in \tau_{A}^Y$ since $U \cap Y \in \tau_{Y}$

Therefore $U \cap A \in \tau_{A}^Y$

Hence: $\tau_{A}^X \subset \tau_{A}^Y$

Therefore: $\tau_{A}^Y=\tau_{A}^X$

$\endgroup$
  • $\begingroup$ When you conclude in the third last line $U \cap A \in \tau ^Y_A$. Just mention before that $(U \cap Y) \cap A =U \cap A$. Otherwise the proof is perfect. $\endgroup$ – Babai Dec 30 '13 at 13:07
  • $\begingroup$ @Susobhan Ok. Thank you $\endgroup$ – user43418 Dec 30 '13 at 13:10
2
$\begingroup$

Let $\tau_{X}$ and $\tau_{Y}$ be the topologies of A as subspace of X and Y.

If $U \in \tau_{X} \Rightarrow \exists B $ open in X such that $U = B \cap A = (B\cap Y) \cap A$ and then $U \in \tau_{Y}$.

If $U \in \tau_{Y} \Rightarrow \exists B $ open in Y such that $U = B \cap A \Rightarrow \exists C $ open in X such that $ B = C \cap Y \Rightarrow U = (C \cap Y)\cap A = C \cap ( Y \cap A) = C \cap A.$

Then $\tau_{X} =\tau_{Y}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.